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If you move your apparatus in that way, then what is a uniform magnetic field in the frame of reference of the copper plate becomes both a magnetic and an electric field in the "laboratory" frame of reference. e.g. See here.

The observed electric field would be $$\vec{E}' = \gamma \vec{v} \times \vec{B}\ ,$$ where $\gamma$ is the usual Lorentz factor.

The observed magnetic field in the laboratory frame would be $$ \vec{B}' = \gamma \vec{B}\ . $$

In the rest frame of the plate, the net force on a charged particle is zero, since the particle would be at rest with respect to the uniform B-field and $\vec{E}=0$.

In the laboratory frame, the force is the sum of that due to $\vec{E}'$ and that due to the motion of the charged particles, moving at $-\vec{v}$ with respect to $\vec{B}'$. Thus $$ \vec{F}' = q(\vec{E}' + (-\vec{v})\times \vec{B'}) = q( \gamma \vec{v} \times \vec{B} - \gamma \vec{v}\times \vec{B}) = 0\ . $$

There would be no separation of the charged particles in either frame of reference.

The "Faraday paradox", where a Lorentz force appears to be induced as if the magnetic field were stationary with respect to a conductor, even when the magnet is moved with the conductor, appears to be confined to rotational motion (e.g., Baumgartel & Maher 2022). It is discussed extensively there, where they point out that a similar phenomenon is not observed for rectilinear motion (the experiment has been done, e.g. Muller 2014). This is fortunate, otherwise Special Relativity would have to be abandoned, since your experiment would offer a way of determining an absolute velocity.

There is no net force on the charged particles and no charge separation. If there were then you would have a way of invalidating Special Relativity because you would have a way of defining an absolute velocity for an inertial frame.

If you move your apparatus in that way, then what is a uniform magnetic field in the frame of reference of the copper plate becomes both a magnetic and an electric field in the "laboratory" frame of reference. e.g. See here.

The observed electric field would be $$\vec{E}' = \gamma \vec{v} \times \vec{B}\ ,$$ where $\gamma$ is the usual Lorentz factor.

The observed magnetic field in the laboratory frame would be $$ \vec{B}' = \gamma \vec{B}\ . $$

In the rest frame of the plate, the net force on a charged particle is zero, since the particle would be at rest with respect to the uniform B-field and $\vec{E}=0$.

In the laboratory frame, the force is the sum of that due to $\vec{E}'$ and that due to the motion of the charged particles, moving at $-\vec{v}$ with respect to $\vec{B}'$. Thus $$ \vec{F}' = q(\vec{E}' + (-\vec{v})\times \vec{B'}) = q( \gamma \vec{v} \times \vec{B} - \gamma \vec{v}\times \vec{B}) = 0\ . $$

There would be no separation of the charged particles in either frame of reference.

The "Faraday paradox", where a Lorentz force appears to be induced as if the magnetic field were stationary with respect to a conductor, even when the magnet is moved with the conductor, appears to be confined to rotational motion (e.g., Baumgartel & Maher 2022). It is discussed extensively there, where they point out that a similar phenomenon is not observed for rectilinear motion (the experiment has been done, e.g. Muller 2014). This is fortunate, otherwise Special Relativity would have to be abandoned, since your experiment would offer a way of determining an absolute velocity. The analysis above cannot be applied to rotating frames since they are not inertial.

If you move your apparatus in that way, then what is a uniform magnetic field in the frame of reference of the copper plate becomes both a magnetic and an electric field in the "laboratory" frame of reference. e.g. See here.

The observed electric field would be $$\vec{E}' = \gamma \vec{v} \times \vec{B}\ ,$$ where $\gamma$ is the usual Lorentz factor.

The observed magnetic field in the laboratory frame would be $$ \vec{B}' = \gamma \vec{B}\ . $$

In the rest frame of the plate, the net force on a charged particle is zero, since the particle would be at rest with respect to the uniform B-field and $\vec{E}=0$.

In the laboratory frame, the force is the sum of that due to $\vec{E}'$ and that due to the motion of the charged particles, moving at $-\vec{v}$ with respect to $\vec{B}'$. Thus $$ \vec{F}' = q(\vec{E}' + (-\vec{v})\times \vec{B'}) = q( \gamma \vec{v} \times \vec{B} - \gamma \vec{v}\times \vec{B}) = 0\ . $$

There would be no separation of the charged particles in either frame of reference.

The "Faraday paradox", where a Lorentz force appears to be induced as if the magnetic field were stationary with respect to a conductor, even when the magnet is moved with the conductor, appears to be confined to rotational motion (e.g., Baumgartle & Maher 2022). It is discussed extensively there, where they point out that a similar phenomenon is not observed for rectilinear motion (e.g., Muller 2014). This is fortunate, otherwise Special Relativity would have to be abandoned, since your experiment would offer a way of determining an absolute velocity.

If you move your apparatus in that way, then what is a uniform magnetic field in the frame of reference of the copper plate becomes both a magnetic and an electric field in the "laboratory" frame of reference. e.g. See here.

The observed electric field would be $$\vec{E}' = \gamma \vec{v} \times \vec{B}\ ,$$ where $\gamma$ is the usual Lorentz factor.

The observed magnetic field in the laboratory frame would be $$ \vec{B}' = \gamma \vec{B}\ . $$

In the rest frame of the plate, the net force on a charged particle is zero, since the particle would be at rest with respect to the uniform B-field and $\vec{E}=0$.

In the laboratory frame, the force is the sum of that due to $\vec{E}'$ and that due to the motion of the charged particles, moving at $-\vec{v}$ with respect to $\vec{B}'$. Thus $$ \vec{F}' = q(\vec{E}' + (-\vec{v})\times \vec{B'}) = q( \gamma \vec{v} \times \vec{B} - \gamma \vec{v}\times \vec{B}) = 0\ . $$

There would be no separation of the charged particles in either frame of reference.

The "Faraday paradox", where a Lorentz force appears to be induced as if the magnetic field were stationary with respect to a conductor, even when the magnet is moved with the conductor, appears to be confined to rotational motion (e.g., Baumgartel & Maher 2022). It is discussed extensively there, where they point out that a similar phenomenon is not observed for rectilinear motion (the experiment has been done, e.g. Muller 2014). This is fortunate, otherwise Special Relativity would have to be abandoned, since your experiment would offer a way of determining an absolute velocity.

If you move your apparatus in that way, then what is a uniform magnetic field in the frame of reference of the copper plate becomes both a magnetic and an electric field in the "laboratory" frame of reference. e.g. See here.

The observed electric field would be $$\vec{E}' = \gamma \vec{v} \times \vec{B}\ ,$$ where $\gamma$ is the usual Lorentz factor.

The observed magnetic field in the laboratory frame would be $$ \vec{B}' = \gamma \vec{B}\ . $$

In the rest frame of the plate, the net force on a charged particle is zero, since the particle would be at rest with respect to the uniform B-field and $\vec{E}=0$.

In the laboratory frame, the force is the sum of that due to $\vec{E}'$ and that due to the motion of the charged particles, moving at $-\vec{v}$ with respect to $\vec{B}'$. Thus $$ \vec{F}' = q(\vec{E}' + (-\vec{v})\times \vec{B'}) = q( \gamma \vec{v} \times \vec{B} - \gamma \vec{v}\times \vec{B}) = 0\ . $$

There would be no separation of the charged particles in either frame of reference.

If you move your apparatus in that way, then what is a uniform magnetic field in the frame of reference of the copper plate becomes both a magnetic and an electric field in the "laboratory" frame of reference. e.g. See here.

The observed electric field would be $$\vec{E}' = \gamma \vec{v} \times \vec{B}\ ,$$ where $\gamma$ is the usual Lorentz factor.

The observed magnetic field in the laboratory frame would be $$ \vec{B}' = \gamma \vec{B}\ . $$

In the rest frame of the plate, the net force on a charged particle is zero, since the particle would be at rest with respect to the uniform B-field and $\vec{E}=0$.

In the laboratory frame, the force is the sum of that due to $\vec{E}'$ and that due to the motion of the charged particles, moving at $-\vec{v}$ with respect to $\vec{B}'$. Thus $$ \vec{F}' = q(\vec{E}' + (-\vec{v})\times \vec{B'}) = q( \gamma \vec{v} \times \vec{B} - \gamma \vec{v}\times \vec{B}) = 0\ . $$

There would be no separation of the charged particles in either frame of reference.

The "Faraday paradox", where a Lorentz force appears to be induced as if the magnetic field were stationary with respect to a conductor, even when the magnet is moved with the conductor, appears to be confined to rotational motion (e.g., Baumgartle & Maher 2022). It is discussed extensively there, where they point out that a similar phenomenon is not observed for rectilinear motion (e.g., Muller 2014). This is fortunate, otherwise Special Relativity would have to be abandoned, since your experiment would offer a way of determining an absolute velocity.